Find limit $\lim_{x \rightarrow 0^+} \frac{x}{x^x-1}$ Find $$\lim_{x \rightarrow 0^+}(x^{x-1}-x^{-1})^{-1}$$
my approach
Firstly I should represent factor in more intuitive form
$$\lim_{x \rightarrow 0^+}\frac{x}{x^x-1} $$
I know that $$\lim_{x \rightarrow 0^+} x^x = 1$$
so I suspect that I have expression of type $ \frac{0}{0} $ 
Ok. Now I am goind to find $$\lim_{x \rightarrow 0^+} \frac{f'(x)}{g'(x)}= \lim_{x \rightarrow 0^+}\frac{1}{(\ln (x) +1)\cdot x^x}$$
Ok but know I have no idea how I can deal with that because $ln x\rightarrow -\infty$ when $x\rightarrow 0^-$ and from other hand $x^x$ is going to $\infty$ and I can't use there Hospital rule again..
 A: $x^x$ does not approach $+\infty$ as $x \to 0^+$. 
The limit $\lim_{x \to 0+} x^x$ is in fact equal to $1$, as we see by expressing $x^x = e^{x\ln{x}}$. 
Thus, your limit $$\lim_{x \to 0+} \frac{1}{(\ln(x) + 1)x^x}$$ 
is in fact equal to $$\lim_{x \to 0+} \frac{1}{\ln(x) + 1} \cdot \lim_{x \to 0+} \frac{1}{x^x},$$ which is equal to $0 \cdot 1 = 0$.
A: $x^x=e^{xln(x)}$, $lim_{x\rightarrow 0}xln(x)=0$. $xln(x)={{ln(x)}\over{1\over x}}$ apply Hospital. You deduce that $lim_{x\rightarrow 0^+}x^x=1$.
A: $$lim_{x \to 0^{+}} \frac{1}{(\ln(x) +1)\cdot x^x} =lim_{x \to 0^{+}} \frac{1}{\ln(x^{x^x}) + \ln(e^{x^x})}  $$ so as $x \to 0$ and $e >0$, $e^{x^x}$$ is approaching infinity faster, therefore the limit goes to zero. 
A: Just another way to do it.
Consider first
$$x^x=e^{x \log(x)}$$ Using Taylor series, then
$$x^x=1+x \log (x)+\frac{1}{2} x^2 \log ^2(x)+O\left(x^3\right)=1+x \log (x)+O\left(x^2\right)$$
$$\frac{x}{x^x-1}=\frac{x}{x \log (x)+O\left(x^2\right) }\sim \frac 1{\log(x)}$$
A: Note that $$\frac{x} {x^x-1}=\frac{x\log x} {\exp(x\log x) - 1}\cdot\frac{1}{\log x} $$ As $x\to 0^{+}$ the expression $\log x\to-\infty $ and $x\log x\to 0$ therefore the first fraction above tends to $1$ and the second fraction above tends to $0$. The desired limit is thus $0$.
